CodeForces 735D - Taxes

题意 给定一个数字n,要求使得找到一个最小的k，满足n=∑ki=1pi，其中pi均为素数。
思路 本来以为这道题目是利用素数的离散性，用类似贪心的思想，每次找到距离n最大的素数ppre，这样能保证n减小的最快。但是实际上这样这样依旧不是最好的。神TM的歌德巴赫猜想。。猜想直接拿来用还叫猜想吗。。不过这个猜想在一定的数据范围内是成立的吧

#include <cstdio>#include <string>#include<iostream>#include<vector>#include <stack>#include <queue>#include <map>#include <cstdlib>#include<string.h>#include <cstring>#include <ctime>#include <algorithm>#include <set>using namespace std;typedef long long ll;typedef pair<int, int>pii;typedef pair<ll, ll> pll;typedef pair<int, ll> pil;typedef vector<vector<ll> >vvi;typedef vector<ll> vi;const int S = 8; //随机算法判定次数，一般8~10就够了 long long pow_mod(long long a, long long n, long long mod){    long long ret = 1;    long long temp = a%mod;    while (n)    {        if (n & 1)ret = ret*temp%mod;        temp = temp*temp%mod;        n >>= 1;    }    return ret;}bool check(long long a, long long n, long long x, long long t){    long long ret = pow_mod(a, x, n);    long long last = ret;    for (int i = 1; i <= t; i++)    {        ret = ret*ret%n;        if (ret == 1 && last != 1 && last != n - 1)return true;//合数          last = ret;    }    if (ret != 1)return true;    else return false;}bool Miller_Rabin(long long n) {    if (n < 2)return false;    if (n == 2)return true;    if ((n & 1) == 0)return false;//偶数      long long x = n - 1;    long long t = 0;    while ((x & 1) == 0)    {        x >>= 1; t++;    }    //srand(time(NULL));/* *************** */    for (int i = 0; i < S; i++)    {        long long a = rand() % (n - 1) + 1;        if (check(a, n, x, t))            return false;    }    return true;}ll getPrePrime(ll n){    while (n)    {        if (Miller_Rabin(n))return n;        n--;    }    return 0;}int main(){    srand(time(NULL));    ll n;    cin >> n;    if (Miller_Rabin(n))printf("1");    else if ((n & 1) == 0)printf("2");    else if (Miller_Rabin(n - 2))printf("2");    else printf("3");    //printf("%d\n", ans);    //system("pause");}